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3 years ago

15

Explain phosphate buffer system

2 answers:

const2013 [10]3 years ago

7 0

Answer:

Phosphate buffer system operates in the internal fluids of all cells. It consists of dihydrogen phosphate ions as the hydrogen ion donor ( acid ) and hydrogen phosphate ion as the ion acceptor ( base ) . ... If extra hydrogen ions enter the cellular fluid then they are neutralised by the hydrogen phosphate ion.

Explanation:

lilavasa [31]3 years ago

6 0

Answer:

Phosphate buffer system operates in the internal fluids of all cells

Explanation:

- Alex :)

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What is the pressure of 0.33 moles of nitrogen gas, if its volume is 15.0 L at –25.0oC?

musickatia [10]

Using the ideal gas law PV =nRTPV=nRT , we find that the pressure will be P =\frac{nRT}{V}P=

V

nRT

. Then, we'll substitute and find the pressure, using T = -25 °C = 248.15 K and R = 0.0821 \frac{atm\cdot L}{mol \cdot K}

mol⋅K

atm⋅L

:

P =\frac{nRT}{V} = \frac{(0.33\,\cancel{mol})(0.0821\frac{atm\cdot \cancel{L}}{\cancel{mol \cdot K}})(248.15\,\cancel{K})}{15.0\,\cancel{L}} = 0.4482\,atmP=

V

nRT

=

15.0

L

(0.33

mol

)(0.0821

mol⋅K

atm⋅

L

)(248.15

K

)

=0.4482atm

In conclusion, the pressure of this gas is P=0.4482 atm.

Reference:

Chang, R. (2010). Chemistry. McGraw-Hill, New York.

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2 years ago

When 1.550 g of liquid hexane (C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87∘C to 38.13∘C.

AlladinOne [14]

Explanation:

1). The given data is as follows.

$T_{i} = 25.87^{o}C$ , $T_{f} = 38.13^{o}C$

C = 5.73 $kJ/^{o}C$

Hence, calculate the change in enthalpy of the reaction as follows.

$\Delta E_{rxn} = -C \times \Delta T$

= $-5.73 \times (38.13 - 25.87)^{o}C$

= -70.25 KJ

As, number of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{1.55}{(6 \times 12 + 14 \times 1)}$

= 0.018 mol

Therefore, enthalpy of reaction in kJ/mol hexane is as follows.

$\Delta E_{rxn} = \frac{-70.25 KJ}{0.018 mol}$

= $-3.90 \times 10^{3}$ kJ/mol

Thus, we can conclude that $\Delta E_{rxn}$ for the reaction in kJ/mol hexane is $-3.90 \times 10^{3}$ kJ/mol .

2). As we know that,

Number of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{1.55}{(7 \times 12 + 8 \times 1)}$

= 0.017 mol

$\Delta E_{rxn} = \E_{rxn} per mol \times \text{number of moles}$

= $-3.91 \times 10^{3} \times 0.017$

= -65.875 kJ

As, $\Delta E_{rxn} = C \times \Delta T
$

-65.875 kJ = $-C \times (37.57 - 23.12)^{o}C$

C = 4.56 $kJ/^{o}C$

Hence, heat capacity of the bomb calorimeter is 4.56 $kJ/^{o}C$ .

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3 years ago

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